Conductor and Conductance and Cell constant Questions in English

(B) The conductivity $(k)$ is related to conductance $(G)$ and cell constant $(l/a)$ by the equation: $k = G \times (l/a)$.
Since conductance $G = 1/R$,we have $k = (1/R) \times (l/a)$.
Rearranging the formula to solve for the cell constant $(l/a)$: $l/a = k \times R$.

(C) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(\kappa)$ is given by the formula:
$\Lambda_m = \frac{1000 \times \kappa}{c}$
Rearranging the formula to solve for conductivity $(\kappa)$:
$\kappa = \frac{\Lambda_m \times c}{1000}$
Given:
$\Lambda_m = 142.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$c = 0.02 \ M = 0.02 \ mol \ L^{-1}$
Substituting the values:
$\kappa = \frac{142.3 \times 0.02}{1000} \ \Omega^{-1} \ cm^{-1}$
$\kappa = 2.846 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
Rounding to two decimal places,we get:
$\kappa \approx 2.85 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$

(A) The conductivity $(\kappa)$ is given by the formula: $\kappa = \frac{\text{cell constant}}{R}$.
Given,cell constant = $0.84 \ cm^{-1}$ and resistance $(R)$ = $14000 \ \Omega$.
Substituting the values: $\kappa = \frac{0.84 \ cm^{-1}}{14000 \ \Omega} = 6.0 \times 10^{-5} \ \Omega^{-1} \ cm^{-1}$.

(C) The dissociation of an $AB_3$ type electrolyte is given by: $AB_3 \rightarrow A^{3+} + 3B^{-}$.
According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution $(\Lambda_m^{\circ})$ is the sum of the molar conductivities of the constituent ions multiplied by their respective stoichiometric coefficients.
Therefore,$\Lambda_m^{\circ} = \nu_{+} \lambda_{+}^{\circ} + \nu_{-} \lambda_{-}^{\circ}$.
Here,$\nu_{+} = 1$ (for $A^{3+}$) and $\nu_{-} = 3$ (for $B^{-}$).
Thus,$\Lambda_m^{\circ} = \lambda_{A^{3+}}^{\circ} + 3 \lambda_{B^{-}}^{\circ}$.

(A) The formula for molar conductivity is $\wedge_{m} = \frac{1000 \times \kappa}{c}$.
Given conductivity $\kappa = 6.065 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$ and concentration $c = 0.005 \ M$.
Substituting the values:
$\wedge_{m} = \frac{1000 \times 6.065 \times 10^{-4}}{0.005} = \frac{0.6065}{0.005} = 121.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(A) The molar conductivity at infinite dilution for an electrolyte is given by Kohlrausch's law: $\Lambda_m^0 = \nu_+ \lambda_+^0 + \nu_- \lambda_-^0$.
For the electrolyte $A_2B_3$,the dissociation reaction is: $A_2B_3 \rightarrow 2A^{3+} + 3B^{2-}$.
Here,the stoichiometric coefficients are $\nu_+ = 2$ and $\nu_- = 3$.
Therefore,the molar conductivity is: $\Lambda_m^0(A_2B_3) = 2 \lambda_{A^{3+}}^0 + 3 \lambda_{B^{2-}}^0$.

(A) According to Kohlrausch's law of independent migration of ions:
$\wedge_m^0(NH_4OH) = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0$
We are given:
$1. \wedge_m^0(Ba(OH)_2) = \lambda_{Ba^{2+}}^0 + 2\lambda_{OH^-}^0 = 520 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$2. \wedge_m^0(BaCl_2) = \lambda_{Ba^{2+}}^0 + 2\lambda_{Cl^-}^0 = 280 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$3. \wedge_m^0(NH_4Cl) = \lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0 = 129 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
To get $\wedge_m^0(NH_4OH)$,we perform the operation: $\frac{1}{2} \wedge_m^0(Ba(OH)_2) + \wedge_m^0(NH_4Cl) - \frac{1}{2} \wedge_m^0(BaCl_2)$
$= \frac{1}{2}(520) + 129 - \frac{1}{2}(280)$
$= 260 + 129 - 140$
$= 249.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

(D) The relationship between conductivity $(k)$,cell constant $(G^*)$,and resistance $(R)$ is given by:
$k = \frac{G^*}{R}$
Given:
$G^* = 0.653 \ cm^{-1}$
$R = 6530 \ \Omega$
Substituting the values:
$k = \frac{0.653 \ cm^{-1}}{6530 \ \Omega} = 1 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$

(B) Molar conductivity $(\Lambda_m)$ is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in a solution.
The formula is $\Lambda_m = \frac{\kappa}{c}$,where $\kappa$ is conductivity $(S \ m^{-1})$ and $c$ is concentration $(mol \ m^{-3})$.
Substituting the units: $\frac{S \ m^{-1}}{mol \ m^{-3}} = S \ m^2 \ mol^{-1}$.
Thus,the $SI$ unit of molar conductivity is $S \ m^2 \ mol^{-1}$.

(A) The cell constant $(G^*)$ of a conductivity cell is determined by measuring the resistance of a standard $KCl$ solution whose conductivity $(\kappa)$ is already known.
Standard solutions typically used for this purpose are $0.01 \ M$,$0.1 \ M$,and $1.0 \ M$ $KCl$ solutions.
$A$ saturated $KCl$ solution is not used because its concentration is extremely high,leading to very high conductivity and potential errors in resistance measurement due to polarization effects and instrument limitations.

(A) The formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{M}$.
Given:
Conductivity $\kappa = 0.0248 \ \Omega^{-1} \ cm^{-1}$
Molarity $M = 0.20 \ M$
Substituting the values:
$\Lambda_{m} = \frac{0.0248 \times 1000}{0.20}$
$\Lambda_{m} = \frac{24.8}{0.20} = 124 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(A) According to Kohlrausch's law of independent migration of ions:
$\wedge^0_{NaBr} = \lambda^0_{Na^+} + \lambda^0_{Br^-}$
$\wedge^0_{NaCl} = \lambda^0_{Na^+} + \lambda^0_{Cl^-} = 126 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\wedge^0_{KBr} = \lambda^0_{K^+} + \lambda^0_{Br^-} = 152 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\wedge^0_{KCl} = \lambda^0_{K^+} + \lambda^0_{Cl^-} = 150 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
To obtain $\wedge^0_{NaBr}$,we perform the operation:
$\wedge^0_{NaBr} = \wedge^0_{NaCl} + \wedge^0_{KBr} - \wedge^0_{KCl}$
$\wedge^0_{NaBr} = 126 + 152 - 150 = 128 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

(B) Kohlrausch's law of independent migration of ions is primarily used to determine the molar conductivity at infinite dilution (zero concentration) for weak electrolytes,such as $NH_4OH$,which do not dissociate completely in solution. For strong electrolytes like $NaNO_3$,$KCl$,and $BaSO_4$,the molar conductivity at zero concentration can be determined by extrapolating the plot of $\Lambda_m$ versus $\sqrt{c}$.

(B) The relationship between conductivity $(\kappa)$,resistance $(R)$,and cell constant $(G^* = \frac{L}{A})$ is given by: $\kappa = \frac{1}{R} \times G^*$.
Given: $\kappa = 1.5 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$ and $R = 150 \ \Omega$.
Substituting the values: $1.5 \times 10^{-4} = \frac{1}{150} \times G^*$.
Therefore,$G^* = 1.5 \times 10^{-4} \times 150 = 0.0225 \ cm^{-1}$.

(A) The cell constant $(G^*)$ is defined as the ratio of the distance between electrodes $(L)$ to the cross-sectional area $(A)$.
Given: $L = 18 \ mm = 1.8 \ cm$ and $A = 2.0 \ cm^2$.
Cell constant $= \frac{L}{A} = \frac{1.8 \ cm}{2.0 \ cm^2} = 0.9 \ cm^{-1}$.

(B) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(K)$ is given by the formula:
$\Lambda_m = \frac{K \times 1000}{M}$
Rearranging the formula to solve for conductivity $(K)$:
$K = \frac{\Lambda_m \times M}{1000}$
Given:
$\Lambda_m = 141 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$M = 0.01 \ M$
Substituting the values:
$K = \frac{141 \times 0.01}{1000} = \frac{1.41}{1000} = 1.41 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$

(A) Conductivity $(\kappa)$ is defined as the reciprocal of resistivity $(\rho)$.
Since the $SI$ unit of resistivity is $\Omega \ m$,the $SI$ unit of conductivity is $\Omega^{-1} \ m^{-1}$.
$\Omega^{-1}$ is also known as Siemens $(S)$.
Therefore,the $SI$ unit of conductivity is $S \ m^{-1}$.

(A) The degree of dissociation $(\alpha)$ is given by the ratio of molar conductivity at a given concentration $(\Lambda_m)$ to the molar conductivity at infinite dilution $(\Lambda_m^{\circ})$.
$\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$
Given,$\Lambda_m = 19.5 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $\Lambda_m^{\circ} = 390 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$\alpha = \frac{19.5}{390} = 0.05$.

(A) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution (zero concentration) can be expressed as:
$\Lambda^{\circ}_{CH_2ClCOOH} = \Lambda^{\circ}_{CH_2ClCOOK} + \Lambda^{\circ}_{HCl} - \Lambda^{\circ}_{KCl}$
Substituting the given values:
$\Lambda^{\circ}_{CH_2ClCOOH} = 1.1 + 4.2 - 1.4$
$\Lambda^{\circ}_{CH_2ClCOOH} = 3.9 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

(D) According to Kohlrausch's law of independent migration of ions:
$\wedge_0(CH_2ClCOOH) = \wedge_0(CH_2ClCOOK) + \wedge_0(HCl) - \wedge_0(KCl)$
Substituting the given values:
$\wedge_0(CH_2ClCOOH) = 1.1 + 4.2 - 1.5$
$\wedge_0(CH_2ClCOOH) = 3.8 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

(A) Given: Molarity $(M) = 0.05 \ M$,Conductivity $(\kappa) = 0.0118 \ S \ cm^{-1}$.
The formula for molar conductivity $(\Lambda_{m})$ is:
$\Lambda_{m} = \frac{\kappa \times 1000}{M}$
Substituting the values:
$\Lambda_{m} = \frac{0.0118 \times 1000}{0.05}$
$\Lambda_{m} = \frac{11.8}{0.05} = 236 \ S \ cm^2 \ mol^{-1}$.

(C) The cell constant is determined by measuring the conductivity of a standard aqueous solution of $KCl$.
This is because the conductivity of $KCl$ solutions is known accurately at various concentrations and different temperatures.

(C) Given: Resistance $(R) = 600 \ \Omega$
Conductivity $(\kappa) = 0.0015 \ \Omega^{-1} \ cm^{-1}$
The formula for cell constant $(G^*)$ is:
$G^* = \kappa \times R$
Substituting the values:
$G^* = 0.0015 \ \Omega^{-1} \ cm^{-1} \times 600 \ \Omega = 0.90 \ cm^{-1}$

(A) The formula for molar conductivity is $\Lambda_m = \frac{1000 \times \kappa}{C}$,where $\kappa$ is the conductivity and $C$ is the molar concentration.
Rearranging for conductivity: $\kappa = \frac{\Lambda_m \times C}{1000}$.
Given: $\Lambda_m = 106 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $C = 0.1 \ M$.
Substituting the values: $\kappa = \frac{106 \times 0.1}{1000} \ \Omega^{-1} \ cm^{-1}$.
$\kappa = \frac{10.6}{1000} \ \Omega^{-1} \ cm^{-1} = 1.06 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$.

(A) Specific conductance or conductivity $(k)$ is defined as the reciprocal of specific resistance or resistivity $(\rho)$.
Mathematically,$k = \frac{1}{\rho}$.
Since resistivity $(\rho)$ is the resistance offered by a unit volume of the conductor,conductivity represents the ease with which current flows through that unit volume.
Therefore,conductivity is inversely proportional to resistivity.

(A) The formula for molar conductivity is $\Lambda_m = \frac{k \times 1000}{M}$.
Given $\Lambda_m = 2.5 \times 10^2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $M = 0.4 \ M$.
Substituting the values: $2.5 \times 10^2 = \frac{k \times 1000}{0.4}$.
$k = \frac{2.5 \times 10^2 \times 0.4}{1000} = \frac{100}{1000} = 0.1 \ \Omega^{-1} \ cm^{-1}$.
Resistivity $\rho$ is the reciprocal of conductivity $k$: $\rho = \frac{1}{k} = \frac{1}{0.1} = 10 \ \Omega \ cm$.

(D) The relationship between molar conductivity $(\Lambda_m)$,conductivity $(k)$,and molarity $(M)$ is given by the formula: $\Lambda_m = \frac{k \times 1000}{M}$.
Given: $\Lambda_m = 230 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $k = 0.0115 \ \Omega^{-1} \ cm^{-1}$.
Substituting the values into the formula: $230 = \frac{0.0115 \times 1000}{M}$.
Rearranging for $M$: $M = \frac{0.0115 \times 1000}{230}$.
$M = \frac{11.5}{230} = 0.05 \ M$.

(B) The relationship between conductivity $(\kappa)$,resistance $(R)$,and cell constant $(G^*)$ is given by:
$\kappa = \frac{1}{R} \times G^*$
Given:
$\kappa = 200 \ \Omega^{-1} \ cm^{-1}$
$G^* = 1 \ cm^{-1}$
Substituting the values:
$200 = \frac{1}{R} \times 1$
$R = \frac{1}{200} \ \Omega$
$R = 0.005 \ \Omega = 5 \times 10^{-3} \ \Omega$

(B) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(k)$ is given by the formula: $\Lambda_m = \frac{k \times 1000}{M}$.
Given: $\Lambda_m = 230 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $M = 0.04 \ M$.
Substituting the values: $230 = \frac{k \times 1000}{0.04}$.
Solving for $k$: $k = \frac{230 \times 0.04}{1000}$.
$k = \frac{9.2}{1000} = 9.2 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.

(C) Kohlrausch's law of independent migration of ions is primarily used to determine the limiting molar conductivity $(\Lambda^0_m)$ of weak electrolytes,which cannot be obtained by extrapolation of the $\Lambda_m$ versus $\sqrt{C}$ plot.
$CH_3COOH$ is a weak electrolyte.
$KCl$,$Na_2SO_4$,and $HCl$ are strong electrolytes whose limiting molar conductivities can be determined directly by extrapolation.

(B) The formula for conductivity $(\kappa)$ is given by: $\kappa = \frac{1}{R} \times \frac{\ell}{A}$
Where $\frac{\ell}{A}$ is the cell constant.
Given: $R = 645 \ \Omega$ and $\text{Cell constant} = 1.29 \ cm^{-1}$.
Substituting the values: $\kappa = \frac{1}{645} \times 1.29$
$\kappa = 0.002 \ \Omega^{-1} \ cm^{-1}$
$\kappa = 2 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$

(C) Specific conductance or Conductivity $(\kappa)$ is defined as the conductance of $1 \ cm^3$ of an electrolytic solution.
Mathematically,$\kappa = \frac{1}{\rho}$,where $\rho$ is the resistivity.
Since $R = \rho \frac{\ell}{A}$,we have $\frac{1}{\rho} = \frac{1}{R} \cdot \frac{\ell}{A}$,which implies $\kappa = G \times G^*$,where $G$ is conductance and $G^*$ is the cell constant.
The $SI$ unit of conductivity is $S \ cm^{-1}$ or $ohm^{-1} \ cm^{-1}$.

(C) The relationship between conductivity $(\kappa)$,resistance $(R)$,and cell constant $(G^*)$ is given by: $\kappa = \frac{1}{R} \times G^*$.
Rearranging for the cell constant: $G^* = \kappa \times R$.
Given: $\kappa = 1.2 \ S \ m^{-1}$ and $R = 30 \ \Omega$.
$G^* = 1.2 \ S \ m^{-1} \times 30 \ \Omega = 36 \ m^{-1}$.
Since $1 \ m = 100 \ cm$,then $1 \ m^{-1} = 10^{-2} \ cm^{-1}$.
$G^* = 36 \times 10^{-2} \ cm^{-1} = 0.36 \ cm^{-1}$.

(D) The formula for molar conductivity is $\Lambda_{m} = \frac{k \times 1000}{M}$.
Given:
Conductivity $(k) = 0.0112 \ \Omega^{-1} \ cm^{-1}$
Molarity $(M) = 0.04 \ M$
Substituting the values:
$\Lambda_{m} = \frac{0.0112 \times 1000}{0.04} = \frac{11.2}{0.04} = 280 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(C) The formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{M}$.
Given:
Conductivity $\kappa = 0.0627 \ S \ cm^{-1}$
Molarity $M = 0.3 \ M$
Substituting the values:
$\Lambda_{m} = \frac{0.0627 \times 1000}{0.3}$
$\Lambda_{m} = \frac{62.7}{0.3} = 209 \ S \ cm^2 \ mol^{-1}$.

(D) The formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{M}$.
Given conductivity $\kappa = 2.67 \times 10^{-4} \ S \ cm^{-1}$ and molarity $M = 0.012 \ M$.
Substituting the values: $\Lambda_{m} = \frac{2.67 \times 10^{-4} \times 1000}{0.012}$.
$\Lambda_{m} = \frac{0.267}{0.012} = 22.25 \ S \ cm^2 \ mol^{-1}$.
Rounding to the nearest provided option,the correct answer is $22.2 \ S \ cm^2 \ mol^{-1}$.

(D) Specific conductance or conductivity $(\kappa)$ is defined as the reciprocal of specific resistance $(\rho)$: $\kappa = \frac{1}{\rho}$.
Since $R = \rho \frac{\ell}{A}$,we have $\frac{1}{\rho} = \frac{1}{R} \times \frac{\ell}{A}$.
Thus,$\kappa = G \times G^*$,where $G$ is conductance and $G^*$ is the cell constant.
Therefore,the conductivity of a solution is defined as the conductance offered by $1 \ cm^3$ of an electrolytic solution.
The $SI$ unit of $\kappa$ is $S \ cm^{-1}$ or $ohm^{-1} \ cm^{-1}$.

(D) The relationship between molar conductivity $(\Lambda_{m})$ and conductivity $(k)$ is given by the formula:
$\Lambda_{m} = \frac{k \times 1000}{M}$
Given:
$\Lambda_{m} = 412.3 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$
$M = 0.02 \ M$
Substituting the values:
$412.3 = \frac{k \times 1000}{0.02}$
Solving for $k$:
$k = \frac{412.3 \times 0.02}{1000}$
$k = 8.246 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$

(A) Given: Conductivity,$\kappa = 1.06 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$ and Concentration,$C = 0.1 \ M$.
The formula for molar conductivity is $\Lambda_m = \frac{\kappa \times 1000}{C}$.
Substituting the values: $\Lambda_m = \frac{1.06 \times 10^{-2} \times 1000}{0.1} = \frac{10.6}{0.1} = 106 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.
Thus,$\Lambda_m = 1.06 \times 10^{2} \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.

(B) Conductivity $(\kappa)$ is the reciprocal of resistivity $(\rho)$.
$\kappa = \frac{1}{\rho}$
Since resistivity $\rho = R \times \frac{A}{l}$,where $R$ is resistance in $\Omega$,$A$ is area in $cm^2$,and $l$ is length in $cm$.
$\kappa = \frac{1}{R} \times \frac{l}{A} = \Omega^{-1} \times \frac{cm}{cm^2} = \Omega^{-1} \ cm^{-1}$.
Thus,the common unit of conductivity in the $C.G.S.$ system is $\Omega^{-1} \ cm^{-1}$ or $S \ cm^{-1}$.

(B) The formula for conductivity $(k)$ is given by: $k = \frac{\text{Cell constant}}{R}$
Given that the cell constant is $0.5 \ cm^{-1}$ and the resistance $(R)$ is $375 \ \Omega$.
Substituting these values into the formula:
$k = \frac{0.5 \ cm^{-1}}{375 \ \Omega} = 1.333 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$

(A) Given: Resistance $(R) = 484 \ \Omega$,Conductivity $(\kappa) = 0.00141 \ \Omega^{-1} \ cm^{-1}$.
Formula for cell constant $(G^*)$ is: $G^* = \kappa \times R$.
Substituting the values: $G^* = 0.00141 \ \Omega^{-1} \ cm^{-1} \times 484 \ \Omega$.
$G^* = 0.68244 \ cm^{-1} \approx 0.682 \ cm^{-1}$.

(C) According to Kohlrausch's Law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.
$\wedge_{m}^{0}(CaCl_{2}) = \lambda_{Ca^{2+}}^{0} + 2 \lambda_{Cl^{-}}^{0}$
Given,$\lambda_{Ca^{2+}}^{0} = 119 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ and $\lambda_{Cl^{-}}^{0} = 71 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.
Substituting the values:
$\wedge_{m}^{0}(CaCl_{2}) = 119 + 2(71)$
$\wedge_{m}^{0}(CaCl_{2}) = 119 + 142 = 261.0 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.

(A) Given: Conductivity $(k)$ = $0.0112 \ \Omega^{-1} cm^{-1}$ and Resistance $(R)$ = $55.0 \ \Omega$.
The formula for cell constant $(G^*)$ is: $G^* = k \times R$.
Substituting the values: $G^* = 0.0112 \ \Omega^{-1} cm^{-1} \times 55.0 \ \Omega$.
Therefore,$G^* = 0.616 \ cm^{-1}$.

(B) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{1000 \times \kappa}{C}$
Here,conductivity $(\kappa)$ = $2 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$ and concentration $(C)$ = $0.08 \ M$.
Substituting the values: $\Lambda_m = \frac{1000 \times 2 \times 10^{-2}}{0.08} = \frac{20}{0.08} = 250 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.

(B) The conductivity $(k)$ of an electrolytic solution is defined as the product of the conductance $(G)$ and the cell constant $(G^*)$.
Since conductance $(G)$ is the reciprocal of electrical resistance $(R)$,we have $G = \frac{1}{R}$.
Substituting this into the formula for conductivity,we get $k = G \cdot G^* = \frac{1}{R} \cdot G^* = \frac{G^*}{R}$.
Therefore,the correct relation is $k = \frac{G^*}{R}$.

(A) Given: Molar conductivity $(\Lambda_m)$ = $124.3 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$,Conductivity $(\kappa)$ = $1.243 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$.
Formula: $\Lambda_m = \frac{1000 \times \kappa}{C}$,where $C$ is the concentration in $mol \ L^{-1}$.
Rearranging for $C$: $C = \frac{1000 \times \kappa}{\Lambda_m}$.
Calculation: $C = \frac{1000 \times 1.243 \times 10^{-4}}{124.3} = \frac{0.1243}{124.3} = 0.001 \ mol \ L^{-1}$.
Therefore,the correct option is $A$.

(D) Conductivity $(k) = \frac{1}{\text{resistivity}} = \frac{1}{2.5 \times 10^{-3} \ \Omega \ cm} = 400 \ \Omega^{-1} \ cm^{-1}$.
Molar conductivity $(\Lambda_{m}) = \frac{1000 \times k}{C}$.
Substituting the values: $\Lambda_{m} = \frac{1000 \times 400}{0.8} = \frac{400000}{0.8} = 5 \times 10^{5} \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.
Thus,the correct option is $D$.

(A) The cell constant $(G^*)$ is defined as the ratio of the distance between the electrodes $(\ell)$ to the area of cross-section $(a)$.
$G^* = \frac{\ell}{a}$
Given: $\ell = 0.98 \ cm$,$a = 1.96 \ cm^{2}$.
$G^* = \frac{0.98 \ cm}{1.96 \ cm^{2}} = 0.5 \ cm^{-1}$.
Therefore,the correct option is $A$.

(D) Given: Concentration $C = 0.01 \ M$,Molar conductivity $\Lambda_m = 400.0 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.
Formula: $\Lambda_m = \frac{1000 \times \kappa}{C}$,where $\kappa$ is the conductivity.
Rearranging for $\kappa$: $\kappa = \frac{\Lambda_m \times C}{1000}$.
Calculation: $\kappa = \frac{400.0 \times 0.01}{1000} = \frac{4}{1000} = 4.0 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.